Given the sequence:
[tex]a_n=\frac{4n}{2n²-3}[/tex]
We evaluate the sequence for n = 12, then:
[tex]\begin{gathered} a_{12}=\frac{4(12)}{2(12)²-3}=\frac{48}{2(144)-3}=\frac{48}{288-3}=\frac{48}{285} \\ \\ \therefore a_{12}=\frac{16}{95} \end{gathered}[/tex]
